Calculating quality factor in a series LCR circuit?

[CrimsonFlash]

Homework Statement

My tutor's notes say that quality factor is supposed to be = 2π (Max. energy stored)/(energy dissipated in one cycle) . So for a standard series LCR circuit, he says:
max energy = 0.5 L I²
energy lost per cycle = 0.5 R I² 2π/ω .
..and at resonance he calculates it to be equal to be [tex] \frac{1}{R} \sqrt{L/C} [\tex]

Homework Equations

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The Attempt at a Solution

However, this is how I tried to do it:
max energy = 0.5 C V²
energy lost per cycle = 0.5 V² /R x 2π/ω
...and at resonance this gives a quality factor of [tex] R \sqrt{C/L} [\tex]

Erm... what da heck?

 

[CrimsonFlash]

Latex...y u no work?

 

[CrimsonFlash]

I'll post the answers here :
Q from the 1st bit: 1/R x (L/C)^(1/2)
Q from my try: R x (C/L)^(1/2)

 

[rude man]

Homework Statement

My tutor's notes say that quality factor is supposed to be = 2π (Max. energy stored)/(energy dissipated in one cycle) . So for a standard series LCR circuit, he says:
max energy = 0.5 L I²
energy lost per cycle = 0.5 R I² 2π/ω .
..and at resonance he calculates it to be equal to be [tex] \frac{1}{R} \sqrt{L/C} [\tex]

Homework Equations

-

The Attempt at a Solution

However, this is how I tried to do it:
max energy = 0.5 C V²
energy lost per cycle = 0.5 V² /R x 2π/ω
...and at resonance this gives a quality factor of [tex] R \sqrt{C/L} [\tex]

Erm... what da heck?

.

Q = wL/R = 1/wRC, w = 1/sqrt(LC)
Q = 2pi(peak energy stored within one cycle)/(energy dissipated during that cycle.)

Now, about your computation: You are using the wrong V_max across the capacitor to compute max stored energy on the capacitor.

max V across capacitor = V0/wRC if the driving voltage is V0 cos(wt).
What is dissipation in R in terms of V0?

Take it from there.

 

[CrimsonFlash]

.

Q = wL/R = 1/wRC, w = 1/sqrt(LC)
Q = 2pi(peak energy stored within one cycle)/(energy dissipated during that cycle.)

Now, about your computation: You are using the wrong V_max across the capacitor to compute max stored energy on the capacitor.

max V across capacitor = V0/wRC if the driving voltage is V0 cos(wt).
What is dissipation in R in terms of V0?

Take it from there.

Max Energy: 0.5 C (V/ωRC)²
Energy lost per cycle: V²/2R x 2π/ω
...and this gives the correct value for Q.

Thank you.